Physics Current Electricity

Given

E₁ = E₂ = E

$I=\frac{2E}{R=r_1+r_2}$   - (i)

Potential drop across second cell is

  $V_A-E_2+I{r}_2=v_B$

According to question V_A – V_B = 0

E₂ -lr₂ = 0

$\Rightarrow R=r_2-r_1$        

R_w = 2R_y

$\Rightarrow \rho \left(\frac{2x}{\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)=\frac{2\rho \left(1-x\right)}{A}$

$\frac{4\rho x}{A}=\frac{2\rho \left(1-x\right)}{A}$

4x = 2 – 2x 6x = 2

x = $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$6$}\right.=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ $\therefore \frac{L_x}{L_y}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{1}{3}*\frac{3}{2}$

$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Based on theoretical data.

J (current density) = 4 * 10⁶ Am^-2

Area between radial distance  $\frac{R}{2}toR$

A =  $\pi \left[R^2-\frac{R^2}{4}\right]=\frac{3R^2}{4}\pi$

I = AJ

=  $\frac{3R^2}{4}\pi *4*10^6$

= 3R² * 10⁶ πAmp

= 3 (4 * 10^-3)² * 10⁶ πAmp

= 3 * 16 * 10^-6 * 10⁶π Amp

= 48πA

R shunt Resistance

Given

CD = 3m

CF = 2m

Current through potentiometer wire

I = $\frac{v}{R_0}=\frac{VA}{\rho L}$ $\frac{}{{}_{}}\frac{}{}$ (1)

A Area of potentiometer wire

$$  $\rho \to$ Resistivity of the wire

L Length of the potentiometer wire

Case 1 When 8 $\Omega$ $$ shunt is bal aced at 3m length

$V_{CD}=I{R}_{CD}=\frac{V}{\rho L}A*\frac{\rho CF}{A}$

$V_{CD}=\frac{V\mathcal{l}}{L}=\frac{V*3}{L}$

VCD = VAB = E – I1r

= $-\frac{E}{R+r}r=\frac{V}{L}*3$ $\frac{}{}\frac{}{}$

$$ $\Rightarrow E\left[1-\frac{r}{8+r}\right]=\frac{V}{L}*3$  (2)

Case 2 When 4 $$ $\Omega$ shunt is balanced at 2m length

$V_{CF}=I{R}_{CF}=\frac{V}{\rho L}A*\frac{\rho CF}{A}=\frac{V*2}{L}$

$V_{CF}=V_{AB}=E-I_{2}r$

= $E-\frac{E}{R+r}r$ $\frac{}{}$

= $E\left[1-\frac{r}{4+r}\right]$ $\frac{}{}$

$\frac{}{}\frac{}{}$  $\Rightarrow E\left(1-\frac{r}{4+r}\right)=\frac{V*2}{L}$ (3)

$\left(2\right)÷\left(3\right)$

$\frac{1-\frac{r}{8+r}}{1-\frac{r}{4+r}}=\frac{3}{2}$

$\Rightarrow \left(\frac{8}{8+r}\right)*\left(\frac{4+8}{4}\right)=\frac{3}{2}$

$\Rightarrow 4\left(4+r\right)=3\left(8+r\right)$

$\Rightarrow 16+4r=24+3r$

r = 24 – 16

r = 8 $\Omega$

$\Rightarrow I_{g}G=\left(l-l_g\right)8\Rightarrow l_g*72=\left(l-l_g\right)8\Rightarrow 9l_g=l-l_g$

$10l_g=l{l}_g=\frac{1}{10}l$

% of I which passes through galvanometer

$$ $l_g=10\mathrm{\%}$  of I

Total power is  $(15*45)+(15*100)+(15*10)+(2*1000)=4325\mathrm{W}$

So, current is $\frac{4325}{220}=19.66\mathrm{A}$

Answer is $20\mathrm{A}\mathrm{m}\mathrm{p}$ .

V_AB = 0

$or,\epsilon -i{r}_1=0$

$\Rightarrow \epsilon -\frac{2\epsilon }{r_1+r_2+R}{r}_1=0\Rightarrow 1=\frac{2r_1}{r_1+r_2+R}$

$\Rightarrow R+r_1+r_2=2r_1$

$R=r_1-r_2$

$H=l^{2}Rt$

$\mathrm{\%}errorinH=\frac{\Delta H}{H}*100=2\left(\frac{\Delta l}{l}\right)*100+\left(\frac{\Delta R}{R}\right)*100+\left(\frac{\Delta t}{t}\right)*100=2*2+1+3=8\mathrm{\%}$

$R_{AB}=5+\left(10//5\right)+10=15+\frac{10*5}{10+5}=15+\frac{10}{3}=\frac{55}{3}k\Omega$

$V_{AB}=R_{AB}l=\left(\frac{55}{3}*10^3\right)*\left(15*10^{-3}\right)=275V$